Combinatorics stars and bars and dearrangement practice

Q1. Natural solutions of  50 < X + y + z <= 100

x + y + z <= 97
x + y + z + w = 97
Total: 100C3

Subtract those when x + y + z <= 50
x + y + z + w = 47
50C3

Answer: 100C3 - 50C3

Q2. Natural solutions of xyz = 2000
Solution:
Factors: 2*2^3*5^3 = 2^4*5^3
It's like distributing 4 apples and 3 oranges among 3 people.
Stars and bars twice:
6C2*5C2 = 150

Q3. Integer solutions of xyz = 3000:
Solution:
Factorize: 2^3.3^1.5^3
Stars and bars thrice:
5C2*3C2*5C2 = 300
This is when all of x,y,z are positive.

Now we can make 2 of  (x,y,z) negative at a time: 3C2
So 4*300 = 1200 is the answer.

Q4. How many ways to put 4 numbered-slips in 4 numbered-boxes so that at least one of them is in the correct box.
Solution:
4C1.D3 + 4C2.D2 + 4C3.D1 + 4C4.D0
 = 4*2 + 6*1 + 4*0 + 1 = 15

Or total - when none is correct = 4! - !4 = 24 - 9 = 15

Q5. 6 letters in 6 envelopes. How may ways if:
a) exactly 2 letters in correct envelopes. 6C2.D4 = 15*9 = 135
b) at least 4 letters go into wrong envelopes:
 6C4.D4 + 6C5.D5 + 6C6.D6
D6 = (6-1)(9+44) = 5*53 = 265
= 135 + 6*44 + 265 = 135 + 264 + 265 = 664

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